My Second Problem

Calculus Level pending

Given that the derivative of $cot(x)$ is $-{csc}^{2}(x)$ ,

first find $\frac{d}{dx}arccot(x)$ .

What is the gradient of the graph $y=arccot(x)$ when $x=2$ ?

1 solution

Daniel Ellesar
Jan 26, 2015

$y = arccotx$

$coty = x$

Implicit differentiation:

$- {csc}^{2}y \frac{dy}{dx} = 1$

$\frac{dy}{dx} = - {sin}^{2}y$

$y = arccotx$

$\frac{dy}{dx} = - sin^{2}(arccotx)$

Draw a triangle.

$\frac{dy}{dx} = - \frac{1}{1+x^{2}}$

$-\frac{1}{1+(2)^{2}} = -0.2$